A) \[\frac{1}{2\pi }\sqrt{\frac{K}{m}}\]
B) \[\frac{1}{2\pi }\sqrt{\frac{({{K}_{1}}+{{K}_{2}})m}{{{K}_{1}}{{K}_{2}}}}\]
C) \[2\pi \sqrt{\frac{K}{m}}\]
D) \[\frac{1}{2\pi }\sqrt{\frac{{{K}_{1}}{{K}_{2}}}{m({{K}_{1}}+{{K}_{2}})}}\]
Correct Answer: D
Solution :
\[n=\frac{1}{2\pi }\sqrt{\frac{{{k}_{eq}}}{m}}=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}{{k}_{2}}}{({{k}_{1}}+{{k}_{2}})m}}\]
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